Strong geodetic number of complete bipartite graphs, crown graphs and hypercubes Valentin Gledel, Vesna Iršič To cite this version: Valentin Gledel, Vesna Iršič. Strong geodetic number of complete bipartite graphs, crown graphs and hypercubes. 2018. hal-01892332 HAL Id: hal-01892332 https://hal.archives-ouvertes.fr/hal-01892332 Preprint submitted on 10 Oct 2018 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Strong geodetic number of complete bipartite graphs, crown graphs and hypercubes Valentin Gledel a Vesna Irˇsiˇc b;c October 9, 2018 a Univ Lyon, Universit´eLyon 1, LIRIS UMR CNRS 5205, F-69621, Lyon, France b Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia c Faculty of Mathematics and Physics, University of Ljubljana, Slovenia Abstract The strong geodetic number, sg(G); of a graph G is the smallest number of vertices such that by fixing one geodesic between each pair of selected vertices, all vertices of the graph are covered. In this paper, the study of the strong geodetic number of complete bipartite graphs is continued. The formula for 0 sg(Kn;m) is given, as well as a formula for the crown graphs Sn. Bounds on sg(Qn) are also discussed. Key words: geodetic number; strong geodetic number; complete bipartite graph; crown graph; hypercube AMS Subj. Class: 05C12, 05C70 1 Introduction The three mainly studied variations of covering vertices of a graph with shortest paths (also called geodesics) are the geodetic problem [4, 3, 6, 7, 9, 14, 15, 21, 27, 29], the isometric path problem [8, 10, 11], and the strong geodetic problem. The latter aims to determine the smallest number of vertices needed, such that by fixing one geodesic between each pair of selected vertices, all vertices of a graph are covered. More formally, the problem is introduced in [24] as follows. Let G = (V; E) be a graph. Given a set S ⊆ V , for each pair of vertices fx; yg ⊆ S, x 6= y, let ge(x; y) be a selected fixed shortest path between x and y. We set Ie(S) = fge(x; y): x; y 2 Sg ; 1 and V (I(S)) = S V (P ). If V (I(S)) = V for some I(S), then the set S is e Pe2Ie(S) e e e called a strong geodetic set. For a graph G with just one vertex, we consider the vertex as its unique strong geodetic set. The strong geodetic problem is to find a minimum strong geodetic set of G. The cardinality of a minimum strong geodetic set is the strong geodetic number of G and is denoted by sg(G). Such a set is also called an optimal strong geodetic set. In the first paper on the topic [24], the strong geodetic number of complete Apollonian networks is determined and it is proved that the problem is NP-complete in general. Also, some comparisons are made with the isometric path problem. The problem has also been studied on grids and cylinders [22], and on Cartesian products in general [18]. Additional results about the problem on Cartesian products, as well as a notion of a strong geodetic core, has been recently studied in [12]. Along with the strong geodetic problem, an edge version of the problem was also introduced in [25]. The strong geodetic problem appears to be difficult even on complete bipartite graphs. Some initial investigation is done in [17], where the problem is presented as an optimization problem and the solution is found for balanced complete bipartite graphs. Some more results have been very recently presented in [19], where it is proved that the problem is NP-complete on general bipartite graphs, but polyno- mial on complete bipartite graphs. The asymptotic behavior of the strong geodetic problem on them is also discussed. In this paper we continue the study on bipartite graphs, specifically on the com- plete bipartite graphs, crown graphs, and hypercubes. In Section 2, we determine the explicit formula for complete bipartite graphs. In Section 3, we discuss the strong geodetic number of crown graphs. In the last section, an upper and lower bound for the strong geodetic number of hypercubes are investigated. To conclude this section we state some basic definitions. Recall that a crown 0 graph Sn is a complete bipartite graph Kn;n without a perfect matching. Recall also n that a hypercube Qn is a graph on the vertex set f0; 1g , where two vertices are adjacent if and only if they differ in exactly one bit. 2 Complete bipartite graphs Strong geodetic number of complete bipartite graphs has been widely studied, as mentioned in Section 1. For completeness we state some already known results. 2 Theorem 2.1 ([17], Theorem 2.1). If n ≥ 6, then p 8 −1 + 8n + 1 >2 ; 8n − 7 is not a perfect square; <> 2 sg(K ) = n;n p > −1 + 8n + 1 >2 − 1; 8n − 7 is a perfect square: : 2 Note also that sg(K1;m) = m for all positive integers m, as K1;m is a tree with m leaves. Hence, in the following results, this case is omitted. To determine sg(Kn;m), we will need the following definitions and notation. We will denote the extension of an integer valued function ' to the real values by 'e. Let 3 ≤ n ≤ m be integers. Define g(p) = m − p for p 2 f0 : : : ; ng and g(p) = m − p 2 e 2 for p 2 R as a continuous extension of g. For p 2 f0; 1; : : : ; ng define f(p) = minfq 2 : q ≥ n − pg and its continuous p Z 2 1+ 1+8(n−p) q extension fe(p) = 2 , the solution of 2 = n − p, where p; q 2 R. Observe that for p < n, f(p) = dfe(p)e, but 0 = f(n) 6= dfe(n)e = 1. Define also F (k) = k + f(k), G(k) = k + g(k) and s(k) = maxfF (k);G(k)g, for all k 2 f0; 1; : : : ; ng. Note that whenever the functions defined above are used, the integers n and m will be clear from the context. Lemma 2.2. If 3 ≤ n ≤ m, then sg(Kn;m) = minfs(k) : 0 ≤ k ≤ n; k 2 Zg. Proof. Let (X; Y ), jXj = n, jY j = m, be the bipartition of Kn;m. Let Sk be a minimal strong geodetic set of the graph, which has exactly k vertices in X. Denote k l = jSk \ Y j. As Y must be covered, l ≥ m − 2 = g(k) (vertices of Y are covered by being in a strong geodetic set or by a geodesic of length two between two vertices in the strong geodetic set in X). l As X must also be covered, l must be such that 2 ≥ n − k. Thus by definition of f, l ≥ f(k). If l ≥ g(k) and l ≥ f(k), then both X and Y are covered. Hence by the minimality of Sk, we have l = maxff(k); g(k)g. Thus sg(Kn;m) = minfjSkj : 0 ≤ k ≤ ng = minfk + maxff(k); g(k)g : 0 ≤ k ≤ ng = minfs(k) : 0 ≤ k ≤ ng : The main idea behind the following result is that sg(Kn;m) is probably close to the value of minfmaxfk + fe(k); k + ge(k)g : 0 ≤ k ≤ ng. But before we state the more general result, consider the case n = 2, which has already been studied in [19, Corollary 2.3]. 3 ( 3; m = 2; Proposition 2.3. If m ≥ 2, then sg(K2;m) = m; m ≥ 3: Theorem 2.4. If 3 ≤ n ≤ m, then 8m; n − 3 ≥ m−3 ; > 2 <> n n m + n − 2 ; m ≥ 2 ; sg(Kn;m) = n n−3 >n; 2 > m ≥ 3 + 2 ; > :min fG(dx∗e − 1);F (dx∗e)g ; otherwise ; p ∗ x 1+ 1+8(n−x) where 3 ≤ x ≤ n − 3 is a solution of m − 2 = 2 . Observe that the first case simplifies to (n; m) 2 f(3; 3); (3; 4); (4; 4); (4; 5); (5; 5); (6; 6)g and that the \otherwise" case appears if and only if m ≥ n ≥ 7 n−3 and m ≤ 3 + 2 . Note that the second case is indeed a known result from [19, Corollary 2.3], but here we present a different proof for it. As the proof consists of some rather technical details, we shall first prove some useful lemmas. Lemma 2.5. For 3 ≤ k ≤ n, the function G(k) is strictly decreasing. Moreover, we have G(0) = G(3) = m and G(1) = G(2) = m + 1. The same holds for the function ge(k). k Proof. It follows from the fact that k − 2 is strictly decreasing for k ≥ 3. Lemma 2.6. For 0 ≤ k ≤ n − 1, F (k) and fe(k) are increasing (not necessarily strictly). Additionally, we have F (n − 1) = n + 1 and F (n) = n. Moreover, jF (k + 1) − F (k)j ≤ 1 for all 0 ≤ k ≤ n − 1. Proof. The claim follows from the fact that if 0 ≤ p ≤ n − 1 and f(p) = q, then f(p + 1) 2 fq; q − 1g by definition of f.
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